5-13 Insert or Merge (25分)

5-13 Insert or Merge   (25分)

According to Wikipedia:

Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.

Merge sort works as follows: Divide the unsorted list into N sublists, each containing 1 element (a list of 1 element is considered sorted). Then repeatedly merge two adjacent sublists to produce new sorted sublists until there is only 1 sublist remaining.

Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer $N$ ($\le 100$). Then in the next line, $N$ integers are given as the initial sequence. The last line contains the partially sorted sequence of the $N$ numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in the first line either "Insertion Sort" or "Merge Sort" to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0

Sample Output 1:

Insertion Sort
1 2 3 5 7 8 9 4 6 0

Sample Input 2:

10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 5 7 4 9 0 6

Sample Output 2:

Merge Sort
1 2 3 8 4 5 7 9 0 6

思路：在Merge函数中参考了算法笔记一书上的非递归实现的解法

令step的初值等于2，然后将数组中每step为一组，将其内部进行排序，之后再将step 乘 2，重复操作直到step/2超过N

#include<stdio.h>  
#include<stdlib.h>  
#include<algorithm>
#define MAX 101
using namespace std;  

bool same(int A[],int change[],int N)
{
	for(int i = 0; i < N; i++)
	{
		if(A[i] != change[i])
		{
			return false;
		}
	}
	return true;
}

void print(int A[],int N)
{
	for(int i = 0; i < N; i++)
	{
		printf("%d",A[i]);
		if(i < N-1)
		printf(" ");
	}
}

bool Insert(int A[],int change[],int N)
{
	int i,Tmp,p;
	bool flag = false;
	for(i = 1; i < N; i++)
	{
		if(i != 1 && same(A,change,N) == true)/*原始序列不能作为比较*/
		{
			flag = true;
		}
		
		Tmp = A[i];
		for(p = i; p > 0 && A[p - 1] > Tmp; p--)
		{
			A[p] = A[p - 1];
		}
		A[p] = Tmp;
		
		if(flag == true)/*这时判断正确的话，说明数组在符合条件的情况下排出了下一步的序列，直接输出数组即可*/
		{
			break;
		}
		
	}
	if(flag == true)
	{
		printf("Insertion Sort\n");
		print(A,N);
		return true;
	}
	else
	{
		return false;
	}
}

void Merge(int A[],int change[],int N)
{
	bool flag = false;
	//step为组内元素个数，step/2是左半部分的元素个数 
	for(int step = 2; step/2 <= N; step *= 2)
	{
		if(step != 2 && same(A,change,N) == true)/*原始序列不能作为比较*/
		{
			flag = true;
		}
		//每step个元素一组，组内[i, min[i + step, N]进行排序 
		for(int i = 0; i < N; i += step)
		{
			sort(A + i, A + min(i + step, N));
		}
		if(flag == true)
		{
		printf("Merge Sort\n");
		print(A,N);
		return;
		}
	}
}

int main(void)
{
	int N;
	scanf("%d",&N);
	int A[MAX];
	int change[MAX];
	int temp[MAX];/*temp用来备份,排序过程在temp数组里进行*/
	int flag = false;
	for(int i = 0; i < N; i++)
	{
		scanf("%d",&A[i]);
		temp[i] = A[i];
	}
	
	for(int i = 0; i < N; i++)
	{
		scanf("%d",&change[i]);
	}
	
	flag = Insert(temp,change,N);
	
	if(flag == false)
	{
		Merge(A,change,N);
	}
	return 0;
}